TITLE: Division of space by hyper-planes
QUESTION [6 upvotes]: It is a well known and lovely result that the maximum number of regions that $\mathbb R^{k}$ (with $k$ positive) can be divided into by $n$ hyperplanes is given by 
$$1+n+\binom{n}{2}+\cdots+\binom{n}{k}.
$$ and occurs when they are in general position. It is clear that the minimum with distinct hyperplanes is $n+1$ (when they are parallel ) and the next smallest is $2n$ when all but one are parallel  and also when each pair has the same (nonempty ) intersection.

For which $n,k,m$ is it possible to divide $\mathbb R^{k}$ into exactly  $m$ regions using $n$ distinct hyperplanes?

Fir purposes of induction it might be better to drop the requirements that the hyperplanes be distinct.

REPLY [2 votes]: The motivation of the question was the observation that the least number of lines in the plane that give five regions is $n=4$ parallel lines. Surprisingly, as long as $m \gt 5$ and $m \leq \binom{n}2+n+1$, the maximum number of regions possible using $n$ lines, there is a division of the plane into $m$ regions using no more than $n$ lines.
I suspected this based on this easy observation (and some computation):

If the plane is divides by $t$ parallel classes of lines with sizes $a_1,\cdots,a_t$, no three lines concurrent, then the plane is split into $\frac{n^2-s}2+n+1$ regions where $n=\sum a_i$ and $s=\sum a_i^2.$

The nice construction by Manturuv does capture some small numbers of regions which this does not, but those attained are enough to establish the result claimed above up to $n=50$ except for $m=5$ and $m=17.$ Naively running over all partitions of $n$ becomes time consuming.
So for $k=2$ then set of attainable number of regions is the union of intervals. Once $n-r \leq \binom{r}2$ we have an interval starting at $$(n-r)(r+1)+\binom{r}{2}-(n-r)=nr-\binom{r+1}2$$ with the previous interval ending at $$(n-r+1)r+\binom{r-1}2=nr-\binom{r+1}2+1.$$ Since this happens near $r=\sqrt{2n},$ we can certainly attain any number of regions  $n\sqrt{2n} \leq m \leq n\frac{n+1}2+1$ using exactly  $n$ lines.